1 ) a . \(3x^2yz-6xyz^2+2xy^2\)
\(=xy\left(3xz-6z^2+2y\right)\)
b . \(x\left(x-y\right)+y\left(y-x\right)\)
\(=x\left(x-y\right)-y\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y\right)\)
\(=\left(x-y\right)^2\)
c . \(x\left(x-y\right)^4-y\left(y-x\right)^3\)
\(=x\left(y-x\right)^4-y\left(y-x\right)^3\)
\(=\left(y-x\right)^3\left[x\left(y-x\right)-y\right]\)
\(=\left(y-x\right)^3\left(xy-x^2-y\right)\)
2 ) \(a.x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b .\(\left(x-2\right)\left(y-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
1 ) a . 3x2yz−6xyz2+2xy23x2yz−6xyz2+2xy2
=xy(3xz−6z2+2y)=xy(3xz−6z2+2y)
b . x(x−y)+y(y−x)x(x−y)+y(y−x)
=x(x−y)−y(x−y)=x(x−y)−y(x−y)
=(x−y)(x−y)=(x−y)(x−y)
=(x−y)2=(x−y)2
c . x(x−y)4−y(y−x)3x(x−y)4−y(y−x)3
=x(y−x)4−y(y−x)3=x(y−x)4−y(y−x)3
=(y−x)3[x(y−x)−y]=(y−x)3[x(y−x)−y]
=(y−x)3(xy−x2−y)=(y−x)3(xy−x2−y)
2 ) a.x2−2x=0a.x2−2x=0
⇔x(x−2)=0⇔x(x−2)=0
⇔[x=0x−2=0⇔[x=0x=2⇔[x=0x−2=0⇔[x=0x=2
Vậy [x=0x=2[x=0x=2
b .(x−2)(y−3)=0(x−2)(y−3)=0
⇔[x−2=0y−3=0⇔[x=2y=3⇔[x−2=0y−3=0⇔[x=2y=3
Vậy [x=2y=3
